So you're asking if the eigenvectors of B determine the eigenvectors of A, given that A and B commute? This doesn't sound right, since the identity matrix commutes with everything. You can narrow down the possible eigenvectors of A, but you won't get a "unique classification."

to see if I understand correctly - let's assume that the matrix A har the eigenvalues {1,2,2} and the matrix B has the eigenvalues {-1,1,1} - then it is possible to construct the eigenvectors of B according to the common unique pairs of A and B( (1,1),(2,1),(2,-1)) giving the following eigenvectors: (1,0,0) , (0,1,1) , (0,-1,1) ?

And had it not been possible with unique pairs of eigenvalues would the eigenvectors not be orthogonanle?